How do I declare an array of undefined or no initial size

Question

I know it could be done using malloc  but I do not know how to use it yet  For example  I wanted the user to input several numbers using an infinite loop with a sentinel to put a stop into it  i e  -1   but since I do not know yet how many he she will input  I have to declare an array with no initial size  but I m also aware that it won t work like this int arr    at compile time since it has to have a definite number of elements  Declaring it with an exaggerated size like int arr 1000   would work but it feels dumb  and waste memory since it would allocate that 1000 integer bytes into the memory  and I would like to know a more elegant way to do this

User · Accepted Answer

This can be done by using a pointer  and allocating memory on the heap using malloc  Note that there is no way to later ask how big that memory block is  You have to keep track of the array size yourself    include  lt stdio h gt   include  lt stdlib h gt   include  lt string h gt   int main int argc  char   argv         declare a pointer do an integer      int  data        we also have to keep track of how big our array is - I use 50 as an example     const int datacount   50    data   malloc sizeof int    datacount      allocate memory for 50 int s      if   data       If data    0 after the call to malloc  allocation failed for some reason        perror  Error allocating memory        abort             at this point  we know that data points to a valid block of memory       Remember  however  that this memory is not initialized in any way -- it contains garbage       Let s start by clearing it       memset data  0  sizeof int  datacount        now our array contains all zeroes       data 0    1    data 2    15    data 49    66     the last element in our array  since we start counting from 0         Loop through the array  printing out the values  mostly zeroes  but even so       for int i   0  i  lt  datacount    i        printf  Element  d   d n   i  data i            That s it  What follows is a more involved explanation of why this works     I don t know how well you know C pointers  but array access in C  like array 2   is actually a shorthand for accessing memory via a pointer  To access the memory pointed to by data  you write  data  This is known as dereferencing the pointer  Since data is of type int    then  data is of type int  Now to an important piece of information   data   2  means  add the byte size of 2 ints to the adress pointed to by data    An array in C is just a sequence of values in adjacent memory  array 1  is just next to array 0   So when we allocate a big block of memory and want to use it as an array  we need an easy way of getting the direct adress to every element inside  Luckily  C lets us use the array notation on pointers as well  data 0  means the same thing as   data 0   namely  access the memory pointed to by data   data 2  means   data 2   and accesses the third int in the memory block

User · Answer

Modern C  aka C99  has variable length arrays  VLA  Unfortunately  not all compilers support this but if yours does this would be an alternative

User · Answer

Try to implement dynamic data structure such as a linked list

User · Answer

The way it s often done is as follows    allocate an array of some initial  fairly small  size  read into this array  keeping track of how many elements you ve read  once the array is full  reallocate it  doubling the size and preserving  i e  copying  the contents  repeat until done    I find that this pattern comes up pretty frequently   What s interesting about this method is that it allows one to insert N elements into an empty array one-by-one in amortized O N  time without knowing N in advance

User · Answer

malloc    and its friends free   and realloc    is the way to do this in C

User · Answer

One way I can imagine is to use a linked list to implement such a scenario  if you need all the numbers entered before the user enters something which indicates the loop termination   posting as the first option  because have never done this for user input  it just seemed to be interesting  Wasteful but artistic   Another way is to do buffered input  Allocate a buffer  fill it  re-allocate  if the loop continues  not elegant  but the most rational for the given use-case    I don t consider the described to be elegant though  Probably  I would change the use-case  the most rational

User · Answer

Here s a sample program that reads stdin into a memory buffer that grows as needed   It s simple enough that it should give some insight in how you might handle this kind of thing   One thing that s would probably be done differently in a real program is how must the array grows in each allocation - I kept it small here to help keep things simpler if you wanted to step through in a debugger   A real program would probably use a much larger allocation increment  often  the allocation size is doubled  but if you re going to do that you should probably  cap  the increment at some reasonable size - it might not make sense to double the allocation when you get into the hundreds of megabytes    Also  I used indexed access to the buffer here as an example  but in a real program I probably wouldn t do that    include  lt stdlib h gt   include  lt stdio h gt    void fatal error void    int main  int argc  char   argv        int buf size   0      int buf used   0       char  buf   NULL      char  tmp   NULL           char c      int i   0       while   c   getchar       EOF            if  buf used    buf size                   need more space in the array               buf size    20               tmp   realloc buf  buf size      get a new larger array              if   tmp  fatal error                  buf   tmp                     buf buf used    c     pointer can be indexed like an array           buf used             puts   n n    Dump of stdin     n         for  i   0  i  lt  buf used    i            putchar buf i               free buf        return 0     void fatal error void        fputs  fatal error - out of memory n   stderr       exit 1       This example combined with examples in other answers should give you an idea of how this kind of thing is handled at a low level

[c] How do I declare an array of undefined or no initial size?

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